Err, why can't a - b + c be equal to -3...? Also, how exactly can it be 19? The max is when a and c are both 9 and b is 0 which gives 18... And, again a - b + 2c could be -4...
correct, that case a - b + c = -3 needs to be loked at. but then a - b + 2c must be even (opppsite parity), so it is at least 10, or -4 (if negative). case 1 (a - b + 2c ≦ 10) => c = (a - b + 2c) - (a - b + c) ≦ 10 - (-3) = 13. contradiction case 2 (a - b + 2c = -4 ) => c = (a - b + 2c) - (a - b + c) = -4 - (-3) = -1. contradiction
Quite so. I find it interesting that the other repies only touched on the negative omissions, whereas the first thing I noticed was that a - b + c cannot be 19.
@hassanakhtar7874 You can do it by hand in less than the runtime of this video. It's not hard, break down into prime factors, group together powers of 10 and 2 (trivial to do mentally) and your left with like 5 multiplications, starting with double digit ones and ending with a 5 digit one. Nothing crazy about doing that by any stretch.
I think it would’ve been good to mention that in the congruence mod 7, we could divide both sides by 2 only because 2 has a multiplicative inverse mod 7. That can’t always be done.
It can be done whenever the number under consideration is relatively prime with the modulus. With a prime modulus, only 0 lacks a multiplicative inverse.
Maybe a simpler solution altogether is to note that 1001=7*11*13 and 99 = 11*9 both divide 14!. Thus we have that the double digit sum have to be 0 mod 99 and the alternating triple digit sum have to be 0 mod 1000. The alternating triple digit sum is 200-bc1+a78-87 = 0 (1000) which gives the delightful equation 113 + a78 -------- = ?bc1 (? can be 0 or 1) which immediately tells us that c = 9 and b=a+1 or b=0 and a=9. The double digit sum gives 7a+bc = 1 (mod 99) which means (after eliminating 199) that 7a+bc = 100, giving a=1, b=2, c=9.
You can simplify the calculations by noticing a-b+c=19--> a+c>=19 and since a,c are positive integers one of a,c is >=10 as 9+9=18 but that would contradict that they are digits.
Reminds me of the problem I was given in the 90s, where given a 10 digit number, with each digit different, the first (left/most significant) is divisible by 1, first two are divisible by 2, first three are divisible by 3. etc. This has a singular answer.
@hassanakhtar7874I think he means the number represented by the first two digits is divisible by 2; the number represented by the first three digits is divisible by 3, and so on.
Some of these cases can be done a bit quicker. For example, here is the missing a - b + c = -3 case a+b+c = 21 a - b + c = -3 then 2a + 2c = 18 => a + c = 9 substitute into first one b = 12 -> impossible a+b+c = 3 a - b + c = -3 2a + 2c = 0 Then c= 0 but this is impossible since c is odd
@@Maths_3.1415 Honestly I think the problem would have been more compelling if we were dealing with some monster like 100!. This 14! is small enough that it's faster to just work it out directly, rather than using all the divisibility tricks.
@@JadeVanadiumResearch Yes, especially as checking the digits given with your computer answer is a good checksum-like way of seeing if you made any computational slip along the way.
It is a nice pattern. If you want to find an even number you look at just the last digit. If you want to determine if divisible by 4, you look at the last 2 digits. If you want to determine if divisible by 8 you look at the last 3 digits, etc. Also you can reduce the number by using some simple rules. say you are div by 8, then you can subtract off 200s without an issue. if you are doing 16, then you can subtract off 2000, etc.
How I would do it - write out 14 factorial, take away the 2, 5 and 10 to make the final two zeros and now I just have to carefully multiply 10 small numbers. I'd do it even without a calculator in 5-6 minutes. And I have a built-in error check having to hit 6 known digits. Also, if you choose your order of multiplication strategically, it would be much easier. Start with the 14, 13 and 12 as these are the two-digits that can make the whole thing gnarly. Multiplying by 11 is easy as you just copy the number shifted by a digit and add them up. Then you have multiplications with single-digit numbers which is simple. That's if we're not allowed calculators...
Even easier if you break things up by prime factors and then do the multiplications in parallel: 14! = 2^11 x 3^5 x 5^2 x 11 x 13 = 100 X 2^9 x 3^5 x 7^2 x 11 x 13 = 100 x 512 x 243 x 49 x 143 = 100 x 512 x 243 x (50 x 143 - 143) = 100 x 512 x 243 x 7007 Which leaves only 2 slightly gnarly multiplications, although even x7007 is easy because of the sole-non-zero repeated digit. But still far quicker to do that the run time of video, especially with the verification method you mentioned.
There is no way that a-b+c can equal 19 when a, b, and c are single-digit integers. The maximum value of the expression is 18. This should have been recognized early on and would have eliminated a lot of the guess-and-check work that came later.
Except that the middle one could also have a-b+c = -3, since -3 ≡ 8 (mod 11). You still have to consider that possibility (although not a-b+c=17, as you rightly stated).
for the final possibility of 21, just add the first 2 equations a+b+c=21 and a-b+c=19 to get 2a+2c=40 or a+c=20 which is impossible with a and c each less than 10
Kind of funny how this is an 18 minute video calculating something in an interesting way that could be calculated with less tools in a boring way in just a few minutes
Perhaps a bit of an overkill solution would be to find 14! mod 10^6 to find the last 6 digits of the number, giving us both b and c. From there, we can use the divisibility rule for 11 to find a.
after a+c+b є {3, 12, 21} a+c-b є {-3, 8, 19} we get 19 is impossible (max 9 - 0 + 9 = 18) and if 8 then a+c+b = 12 (both must be even) So b=2, a+c=10 and mod 13 gives 0=87-a78+bc1-200=100b+10c-100a-190 => c+3a-3b є {-19, -7, 6, 19, 42} but c+3a-3b = 2a+4, so 2a+4 = 6 => a=1, b=2, c=0. if -3 then a+c+b=3 (both must be odd) => b=3, a+c=0, so c = 0, but mod 32 c must be odd.
If a-b+c is at least 8 then a+b+c cannot be 3. Furthermore, the difference between a+b+c and a-b+c is 2b, so they also have the same polarity. Clearly, b=1 or b=2. If b were 1, then a+c = 20. Thus b=2 and a+c=10.
You'll find that a-b+c could be -3 (because -3 ≡ 8 mod 11), so it's not at least 8 and you can't automatically disqualify a+b+c=3 simply because it's less than 8. However, if a-b+c = -3 and a+b+c = 3, then you have 2a + 2c = 0, making a and c both zero. But we know c is odd, and _that_ disqualifies the a-b+c = -3 case and consequently the a+b+c = 3 case.
The difference between a+b+c and a-b+c is 2b. The difference is therefore a multiple of 2. If two numbers have a difference of a multiple of 2, then either both numbers are odd, or both numbers are even. That means they have the same parity (the parity of a number means whether it's even or odd).
We know that the number on the right hand side is divisible by 9 because it is equal to 14! on the left hand side - and 14! is obviously divisible by 9. And a number is divisible by 9 _precisely_ when the sum of its digits is divisible by 9. As far as I know, usually one learns that in elementary school...?
Because 9 divides 14! and since it is written in decimal system, i.e. in base 10, the sum of digits must be divisible by 9. Why? It's simple - every number is representet as follows: a = a(n) * 10^n + a(n-1) * 10^(n-1) + ... + a(1) * 10 + a(0) Divisibility by 9 here translates to 1 = (10^i mod 9). So, every power of 10 here reduces to 1, therefore we must consider the sum of the digits only.
This works for all numbers such that it is one less then the base. so if you are working base 8, then 7 works like that. If you are working base 16, then 15 would work like that. The alternating on 11 works the same way because it is the base plus 1.
@@ingiford175 ... and hence (for completeness) every power of the base is congruent to (-1) raised to that power, which means we apply alternating signs when calculating the digit sum.
Err, why can't a - b + c be equal to -3...? Also, how exactly can it be 19? The max is when a and c are both 9 and b is 0 which gives 18...
And, again a - b + 2c could be -4...
The case where a-b+c=-3 is forgoten
correct, that case a - b + c = -3 needs to be loked at. but
then a - b + 2c must be even (opppsite parity), so it is at least 10, or -4 (if negative).
case 1 (a - b + 2c ≦ 10) => c = (a - b + 2c) - (a - b + c) ≦ 10 - (-3) = 13. contradiction
case 2 (a - b + 2c = -4 ) => c = (a - b + 2c) - (a - b + c) = -4 - (-3) = -1. contradiction
@@christophniessl9279Indeed, I'm not saying it gives another possible solution, just that he didn't show that it doesn't...
@@ivanlazaro7444 Came here to say he missed a case. Seems comments got it covered.
Quite so. I find it interesting that the other repies only touched on the negative omissions, whereas the first thing I noticed was that a - b + c cannot be 19.
I think I will just calculate 14!, since I have already known that it only has 11 digits.
I threw it into Matlab and it had the answer, but that's more engineering than entertaining!
@hassanakhtar7874 You can do it by hand in less than the runtime of this video. It's not hard, break down into prime factors, group together powers of 10 and 2 (trivial to do mentally) and your left with like 5 multiplications, starting with double digit ones and ending with a 5 digit one. Nothing crazy about doing that by any stretch.
@ 5:14 a-b+c = -3 is also a potential solution in the evaluation modulo 11. Similarly, a-b+2c = -4 should a possibility in modulo 7 evaluation.
I think it would’ve been good to mention that in the congruence mod 7, we could divide both sides by 2 only because 2 has a multiplicative inverse mod 7. That can’t always be done.
It can be done whenever the number under consideration is relatively prime with the modulus. With a prime modulus, only 0 lacks a multiplicative inverse.
@@frenchguy7518 making *Z* / p *Z* a field :D
@@schweinmachtbree1013 Exactly.
Maybe a simpler solution altogether is to note that 1001=7*11*13 and 99 = 11*9 both divide 14!. Thus we have that the double digit sum have to be 0 mod 99 and the alternating triple digit sum have to be 0 mod 1000. The alternating triple digit sum is 200-bc1+a78-87 = 0 (1000) which gives the delightful equation
113
+ a78
--------
= ?bc1
(? can be 0 or 1)
which immediately tells us that c = 9 and b=a+1 or b=0 and a=9. The double digit sum gives 7a+bc = 1 (mod 99) which means (after eliminating 199) that 7a+bc = 100, giving a=1, b=2, c=9.
99 = 11*7? Am I missing something?
@@WhatIKnowAboutThat 99=11*9, typo corrected.
That was the first way I thought of doing it. Cover all of the factors, not just 11.
Best answer
The method is correct, but the calculation is wrong. It should be mod (1001) not mod (1000). And it will lead to only one condition ?=0
You can simplify the calculations by noticing a-b+c=19--> a+c>=19 and since a,c are positive integers one of a,c is >=10 as 9+9=18 but that would contradict that they are digits.
Reminds me of the problem I was given in the 90s, where given a 10 digit number, with each digit different, the first (left/most significant) is divisible by 1, first two are divisible by 2, first three are divisible by 3. etc. This has a singular answer.
@hassanakhtar7874I think he means the number represented by the first two digits is divisible by 2; the number represented by the first three digits is divisible by 3, and so on.
@hassanakhtar7874 The number 1234567890 works until you get to the 4th place. 1 | 1, 2 | 12, 3 | 123, 4 does not divide 1234, etc.
Subtract (a-b+c) from a+b+c and you eliminate the odd parity possibility and find b = 2 before considering the 3rd equation.
How can a-b+c equal 19? Am I missing something? One of the a or c would need to be 10 for that to be possible?
You missed nothing. a-b+c=19 is simply not a valid case
Some of these cases can be done a bit quicker. For example, here is the missing a - b + c = -3 case
a+b+c = 21
a - b + c = -3
then 2a + 2c = 18 => a + c = 9
substitute into first one
b = 12 -> impossible
a+b+c = 3
a - b + c = -3
2a + 2c = 0
Then c= 0 but this is impossible since c is odd
17:34 Real mathematicians calculate 14! by hand and don’t use these cheap tricks 😎
Yes even if it is 100!
@@Maths_3.1415 Honestly I think the problem would have been more compelling if we were dealing with some monster like 100!. This 14! is small enough that it's faster to just work it out directly, rather than using all the divisibility tricks.
@@JadeVanadiumResearch Yes, especially as checking the digits given with your computer answer is a good checksum-like way of seeing if you made any computational slip along the way.
That 32 was a trick I have not heard of:)
It is a nice pattern. If you want to find an even number you look at just the last digit. If you want to determine if divisible by 4, you look at the last 2 digits. If you want to determine if divisible by 8 you look at the last 3 digits, etc. Also you can reduce the number by using some simple rules. say you are div by 8, then you can subtract off 200s without an issue. if you are doing 16, then you can subtract off 2000, etc.
@@ingiford175 Thanks for info, much appreciated
How I would do it - write out 14 factorial, take away the 2, 5 and 10 to make the final two zeros and now I just have to carefully multiply 10 small numbers. I'd do it even without a calculator in 5-6 minutes. And I have a built-in error check having to hit 6 known digits. Also, if you choose your order of multiplication strategically, it would be much easier. Start with the 14, 13 and 12 as these are the two-digits that can make the whole thing gnarly. Multiplying by 11 is easy as you just copy the number shifted by a digit and add them up. Then you have multiplications with single-digit numbers which is simple.
That's if we're not allowed calculators...
Even easier if you break things up by prime factors and then do the multiplications in parallel:
14! = 2^11 x 3^5 x 5^2 x 11 x 13
= 100 X 2^9 x 3^5 x 7^2 x 11 x 13
= 100 x 512 x 243 x 49 x 143
= 100 x 512 x 243 x (50 x 143 - 143)
= 100 x 512 x 243 x 7007
Which leaves only 2 slightly gnarly multiplications, although even x7007 is easy because of the sole-non-zero repeated digit. But still far quicker to do that the run time of video, especially with the verification method you mentioned.
If brute force, that's smart brute force. But there's a very smart solution mod 1001 and mod 99
7:43 I honestly thought he would say 64, from which we would be able to conclude a fact about b’s parity given that we already know c must be odd.
There is no way that a-b+c can equal 19 when a, b, and c are single-digit integers. The maximum value of the expression is 18. This should have been recognized early on and would have eliminated a lot of the guess-and-check work that came later.
The ending could have gone quite a bit faster. a-b+c
Except that the middle one could also have a-b+c = -3, since -3 ≡ 8 (mod 11). You still have to consider that possibility (although not a-b+c=17, as you rightly stated).
I think we can also eliminate a-b+2c=-4, because it is obviously a-b+c
Let m ethink - if a-b+c has to be equal 19, that means that a or c has to be equal to 10, another one 9, and b=0, but 10 is not a digid...
Thanks for the wonderful problem!
for the final possibility of 21, just add the first 2 equations a+b+c=21 and a-b+c=19 to get 2a+2c=40 or a+c=20 which is impossible with a and c each less than 10
It is interesting, but in this particular case it would be easier to calculate 14!.
5:02 what about -3? Also 19 isn't possible even if b is 0, 9+9-0=18.
in a test i would brute force the answer!
Kind of funny how this is an 18 minute video calculating something in an interesting way that could be calculated with less tools in a boring way in just a few minutes
Perhaps a bit of an overkill solution would be to find 14! mod 10^6 to find the last 6 digits of the number, giving us both b and c. From there, we can use the divisibility rule for 11 to find a.
after a+c+b є {3, 12, 21} a+c-b є {-3, 8, 19} we get 19 is impossible (max 9 - 0 + 9 = 18) and
if 8 then a+c+b = 12 (both must be even) So b=2, a+c=10 and
mod 13 gives 0=87-a78+bc1-200=100b+10c-100a-190 => c+3a-3b є {-19, -7, 6, 19, 42} but c+3a-3b = 2a+4, so 2a+4 = 6 =>
a=1, b=2, c=0.
if -3 then a+c+b=3 (both must be odd) => b=3, a+c=0, so c = 0, but mod 32 c must be odd.
why pick these particular numbers (9, 11, 32) to divide by?
Why we were looking for other solutions, 14! can't have multiple decimal representations)
If a-b+c is at least 8 then a+b+c cannot be 3.
Furthermore, the difference between a+b+c and a-b+c is 2b, so they also have the same polarity. Clearly, b=1 or b=2.
If b were 1, then a+c = 20. Thus b=2 and a+c=10.
You'll find that a-b+c could be -3 (because -3 ≡ 8 mod 11), so it's not at least 8 and you can't automatically disqualify a+b+c=3 simply because it's less than 8.
However, if a-b+c = -3 and a+b+c = 3, then you have 2a + 2c = 0, making a and c both zero. But we know c is odd, and _that_ disqualifies the a-b+c = -3 case and consequently the a+b+c = 3 case.
The alternating chucks (of 3 digits) works for 7, 11, and 13 because 7*11*13 = 1001
mfs who calculated it by hand:
Naturally 14! Will be divisible by 7
Not clear to me why the first two equations have the same parity. Someone can explain again?
The difference between a+b+c and a-b+c is 2b. The difference is therefore a multiple of 2. If two numbers have a difference of a multiple of 2, then either both numbers are odd, or both numbers are even. That means they have the same parity (the parity of a number means whether it's even or odd).
why does adding the digits make it divisble by 9 ??
We know that the number on the right hand side is divisible by 9 because it is equal to 14! on the left hand side - and 14! is obviously divisible by 9.
And a number is divisible by 9 _precisely_ when the sum of its digits is divisible by 9. As far as I know, usually one learns that in elementary school...?
Because 9 divides 14! and since it is written in decimal system, i.e. in base 10, the sum of digits must be divisible by 9.
Why? It's simple - every number is representet as follows: a = a(n) * 10^n + a(n-1) * 10^(n-1) + ... + a(1) * 10 + a(0)
Divisibility by 9 here translates to 1 = (10^i mod 9). So, every power of 10 here reduces to 1, therefore we must consider the sum of the digits only.
This works for all numbers such that it is one less then the base. so if you are working base 8, then 7 works like that. If you are working base 16, then 15 would work like that. The alternating on 11 works the same way because it is the base plus 1.
@@ingiford175 ... and hence (for completeness) every power of the base is congruent to (-1) raised to that power, which means we apply alternating signs when calculating the digit sum.
pour a-b+c appartient à 8 seulement car a=9b=0et c=9 total 18
How is a - b + c = 19 ? And how can it NOT be -3 ? Should be a - b + c € { -3, 8 }
I mean, I have up to 10! memorized, so just that x 132 x 182
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